Question

British and American spelling are `rigour' and `rigor', respectively. A man staying at a Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40% of the English-speaking men at the hotel are British and 60% are Americans, what is the probability that the writer is British?

Answer 1

45.45%

Explanation:

Given:

British spelling as "rigour"

here Probability of selecting vowel , P(V/B) =
`\frac{\textup{Number of vowels}}{\textup{Total letters}}`

or

Probability of selecting vowel , P(V/B) =
`\frac{\textup{3}}{\textup{6}}` = 0.5

and,

American spelling as "rigor"

here Probability of selecting vowel , P(V/A) =
`\frac{\textup{Number of vowels}}{\textup{Total letters}}`

or

Probability of selecting vowel , P(V/A) =
`\frac{\textup{2}}{\textup{5}}` = 0.4

Probability of English-speaking men at hotel are British P(B) = 40% = 0.4

Probability of English-speaking men at hotel are American P(A) = 60% = 0.6

Now using Baye's theorem

the probability that the writer is British P(B/V) =
`(P(B)* P(V/E))/(P(B)* P(V/E)+P(A)* P(V/A))`

or

the probability that the writer is British P(B/V) =
`(0.4*0.5)/(0.4*0.5+0.6*0.4)`

or

the probability that the writer is British P(B/V) =
`(0.2)/(0.44)`

or

the probability that the writer is British P(B/V) = 0.4545 = 45.45%